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# If n and y are positive integers and 450y=n^3, which of the

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Joined: 21 Nov 2007
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If n and y are positive integers and 450y=n^3, which of the [#permalink]

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27 Nov 2007, 11:30
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If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)

Kudos [?]: [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3585

Kudos [?]: 4478 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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27 Nov 2007, 12:55
only I

450=3*5*3*2*5=2*3^2*5^2

y=k*2^2*3*5 where k - positive integer

Kudos [?]: 4478 [0], given: 360

SVP
Joined: 05 Jul 2006
Posts: 1742

Kudos [?]: 418 [0], given: 49

Re: PS - integer [#permalink]

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27 Nov 2007, 12:56
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)

450Y = 5^2* 2*3^2 Y = N^3 , THEN Y MUST HAVE AT LEAST A FACTOR OF 5 AND 2^2 AND A 3

A is my answer

Kudos [?]: 418 [0], given: 49

CEO
Joined: 29 Mar 2007
Posts: 2554

Kudos [?]: 500 [0], given: 0

Re: PS - integer [#permalink]

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27 Nov 2007, 15:25
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)

A by testing the answers.

since n and y are + integers. n^3 is a perfect cube. So find the answer choice that when multiplied by 450 is a perfect cube.

Only I works: 3*4*5=60 --> 6*45=270 Then just add 2 zeros.

II and III do not work.

Kudos [?]: 500 [0], given: 0

SVP
Joined: 28 Dec 2005
Posts: 1549

Kudos [?]: 173 [0], given: 2

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27 Nov 2007, 19:23
walker wrote:
only I

450=3*5*3*2*5=2*3^2*5^2

y=k*2^2*3*5 where k - positive integer

can someone explain the logic here ? i mean, i can break 450 down into primes, but how do we determine that 1 is the right answer ?

Kudos [?]: 173 [0], given: 2

CEO
Joined: 17 Nov 2007
Posts: 3585

Kudos [?]: 4478 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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27 Nov 2007, 21:10
pmenon wrote:
can someone explain the logic here ? i mean, i can break 450 down into primes, but how do we determine that 1 is the right answer ?

for example II:

z= y / (9*2*5) = k*2^2*3*5 / (3*3*2*5)=k*2/3 ==> k=3: z-integer, but k=4: z-fraction. Therefore, II is insuff

Kudos [?]: 4478 [0], given: 360

Director
Joined: 09 Aug 2006
Posts: 754

Kudos [?]: 239 [0], given: 0

Re: PS - integer [#permalink]

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27 Nov 2007, 22:15
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)

n^3 = 450 y
n^3 = 2 * 3^2 * 5^2 * y
Since n is an integer, the cube root of the right side of above equation should be an integer as well. The simplest way to make it an integer is if y = 2^2 * 3 * 5 = 60.

Now divide 60 by I II & III to see which work. Only I does so answer is A.

Kudos [?]: 239 [0], given: 0

SVP
Joined: 28 Dec 2005
Posts: 1549

Kudos [?]: 173 [0], given: 2

Re: PS - integer [#permalink]

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16 Dec 2007, 10:18
GK_Gmat wrote:
tennisgurun wrote:
If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y / (3*4*5)
II. y / (9*2*5)
III. y / (3*2*25)

n^3 = 450 y
n^3 = 2 * 3^2 * 5^2 * y
Since n is an integer, the cube root of the right side of above equation should be an integer as well. The simplest way to make it an integer is if y = 2^2 * 3 * 5 = 60.

Now divide 60 by I II & III to see which work. Only I does so answer is A.

i like this approach very much. thanks a lot for enlightening me

Kudos [?]: 173 [0], given: 2

Re: PS - integer   [#permalink] 16 Dec 2007, 10:18
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# If n and y are positive integers and 450y=n^3, which of the

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